Worked examples — Use Thevenin equivalent circuits

Before we start, one reminder in plain words. A circuit has two terminals — think of them as the two metal prongs you'd touch a meter to. Everything to the left of those prongs is "the network"; everything to the right is "the load." Thevenin says the whole left side, no matter how tangled, acts like one battery $V_{th}$ behind one resistor $R_{th}$.

• $V_{th}$ = voltage across the prongs with nothing attached (open circuit).
• $R_{th}$ = the "stiffness" you feel looking into the prongs once every internal battery/current-pump is switched off.

Everything below is a rehearsal of finding those two numbers in every awkward situation.

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The scenario matrix

Each row is a case class — a genuinely different kind of circuit. Every one gets at least one fully worked example below (the links jump to it).

#	Case class	What's tricky about it	Example
A	Single voltage source + divider	The baseline (from parent)	Ex 1
B	Two voltage sources fighting	Signs — which way does current go?	Ex 2
C	Current source in the network	I-source → open when deactivated	Ex 3
D	Degenerate load: short & open	$R_L=0$ and $R_L=\infty$ limits	Ex 4
E	Dependent source present	Can't deactivate — must use test source	Ex 5
F	$V_{th}=0$ (balanced bridge)	A network that looks live but is dead	Ex 6
G	Real-world word problem	Battery + wiring, model a real thing	Ex 7
H	Exam twist: given measurements	Work backwards from two meter readings	Ex 8

Recall The two escape routes for

$R_{th}$ No dependent source? ::: Deactivate independent sources (V→short, I→open), find the resistance looking in. Dependent source present? ::: Apply a test source $V_{test}$, measure $I_{test}$, take $R_{th}=V_{test}/I_{test}$. Either way, an alternative: ::: $R_{th}=V_{th}/I_{sc}$ using the short-circuit current.

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Example 1 — Cell A: the baseline single source

Forecast: guess whether $V_{th}$ is above or below 6 V, and whether $R_{th}$ is above or below the smaller resistor $R_2$.

Reading the figure: on the left you see the actual circuit — the blue vertical bar is the 12 V source, the orange zig-zag along the top is $R_1$, the blue vertical zig-zag is $R_2$, and the two red dots are the terminals. On the right is the redraw after we switch the source off: the source becomes a plain wire, so both resistors (orange $R_1$, blue $R_2$) now dangle from the same terminal node down to the same ground rail — the green arrow points at that parallel pair. Keep both pictures in view; the left gives $V_{th}$, the right gives $R_{th}$.

Step 1 — open-circuit voltage. With the load gone, no current is pulled out of the terminal, so $R_1$ and $R_2$ carry the same current in a plain voltage divider: $V_{th}=V_S\frac{R_2}{R_1+R_2}=12\cdot \frac{2}{6}=4\text{ V}$ Why this step? Open circuit means zero terminal current, which is the exact condition a divider formula assumes.

Step 2 — Thevenin resistance. Switch $V_S$ off (replace it with a wire). Now the top of $R_1$ is tied to ground, so both $R_1$ and $R_2$ hang from the terminal node down to ground — that's parallel: $R_{th}=R_1\Vert R_2=\frac{4\cdot 2}{4+2}=1.33\Omega$ Why this step? Deactivating the only source leaves pure resistance; the right-hand redraw (green arrow) shows the two resistors sharing both endpoints — the visual signature of "in parallel."

Step 3 — reattach the load. Now it's just $V_{th}$ pushing through $R_{th}+R_L$ in series: $I_L=\frac{V_{th}}{R_{th}+R_L}=\frac{4}{1.33+6}=0.545\text{ A}$ Why this step? Thevenin's whole payoff: once the tangle is reduced to $V_{th}$ behind $R_{th}$, the load simply adds one more series resistor, so the current is a single Ohm's-law division — no divider or nodal work needed anymore.

Verify: terminal voltage $V=I_L{R}_L=0.545\cdot 6=3.27\text{ V}$. Plug into the terminal law: $V_{th}-I_L{R}_{th}=4-0.545\cdot 1.33=3.27\text{ V}$ ✅ — the two ways of computing the terminal voltage agree, so the equivalent is self-consistent. Forecast answered: $V_{th}=4\text{ V}$ is below 6 V, and $R_{th}=1.33\Omega$ is below $R_2=2\Omega$ (parallel is always smaller than either resistor).

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Example 2 — Cell B: two sources fighting

Forecast: the stronger source is $V_A$. Will the node sit closer to 10 V or to 4 V, and can it be outside the 4–10 V range?

Reading the figure: the big red dot in the centre is the shared node (this is $V_{th}$'s home). Coming in from the upper-left is the blue 10 V source through the orange resistor $R_A$; coming in from the upper-right is the green 4 V source through the blue resistor $R_B$. The grey horizontal line at the bottom is the common ground both sources return to. The red arrow at the bottom reminds you which way current actually flows — out of the stronger 10 V branch, across the node, and into the weaker 4 V branch.

Step 1 — open-circuit node voltage. With nothing drawn out of the node, whatever current leaves $V_A$'s branch must enter $V_B$'s branch. Use a nodal balance at the node voltage $V$: $\frac{V-V_A}{R_A}+\frac{V-V_B}{R_B}=0$ Why this step? Kirchhoff's current law: the currents into the node must sum to zero, because there's no load to soak up any.

Step 2 — solve for $V$. Both denominators are $5$. To get rid of the fractions we multiply every term by $5$ (multiplying both sides of an equation by the same number keeps it true); each $\frac{\cdots }{5}$ then becomes just its numerator: $5\cdot \frac{V-10}{5}+5\cdot \frac{V-4}{5}=5\cdot 0\Rightarrow (V-10)+(V-4)=0$ $\Rightarrow 2V=14\Rightarrow V_{th}=7\text{ V}$ Why this step? Clearing the denominators turns a fraction equation into plain arithmetic. Equal resistors make the node the plain average of the two source voltages — it lands between them, never outside. Sign sanity: it's pulled toward the bigger source $V_A$, which is exactly where 7 V leans.

Step 3 — Thevenin resistance. Deactivate both voltage sources (short each). Now $R_A$ and $R_B$ both run from the node to ground → parallel: $R_{th}=R_A\Vert R_B=\frac{5\cdot 5}{10}=2.5\Omega$ Why this step? Once shorted, the two source-tops both sit at ground, so the resistors share both endpoints.

Verify: short-circuit current route. Short the node to ground: $I_{sc}=\frac{V_A}{R_A}+\frac{V_B}{R_B}=\frac{10}{5}+\frac{4}{5}=2.8\text{ A}$. Then $R_{th}=V_{th}/I_{sc}=7/2.8=2.5\Omega$ ✅. Two methods agree. Forecast answered: the node lands at 7 V — closer to the stronger 10 V source and safely inside the 4–10 V range (a passive resistive network can never push the node past either source).

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Example 3 — Cell C: a current source inside

Forecast: a current source deactivates by becoming an open gap (opposite of a voltage source). Guess: after deactivation, does current still flow through $R_1$?

Reading the figure: the blue circle on the left with the upward arrow inside is the current source — the arrow shows it pumping 3 A upward, into the node. The red dot labelled "node" is where that current arrives. From the node, the orange vertical zig-zag $R_1$ drops to the grey ground rail, and the green horizontal zig-zag $R_2$ heads right to the terminal (red dots). The dashed grey line between the terminals is the open gap — nothing bridges them, so no current can leave through $R_2$. That dashed gap is the whole reason $R_2$ carries zero current in Step 1.

Step 1 — open-circuit voltage. Terminals open ⇒ no current through $R_2$ ⇒ no voltage drop across $R_2$. So the terminal voltage equals the node voltage. All 3 A from the source must flow through $R_1$ (nowhere else to go): $V_{node}=I_S\cdot R_1=3\cdot 6=18\text{ V}\Rightarrow V_{th}=18\text{ V}$ Why this step? An open terminal forces the $R_2$ current to zero, so $R_2$ is "invisible" for $V_{th}$ — its drop is $I{R}_2=0$.

Step 2 — Thevenin resistance. Deactivate the current source: replace it with an open gap (a current source turned off pushes zero current, which is what an open wire does). Now trace from the terminal: through $R_2$, into the node, then $R_1$ to ground — a clean series path: $R_{th}=R_1+R_2=6+3=9\Omega$ Why this step? With the source gapped, the branch that used to inject current is gone, so $R_1$ and $R_2$ are simply nose-to-tail.

Verify: short-circuit current. Short the terminals; the node-to-ground now has $R_1$ and (through the short) $R_2$ in parallel, and $I_S$ splits by current divider. Current through $R_2$ (the short branch) is $I_{sc}=I_S\frac{R_1}{R_1+R_2}=3\cdot \frac{6}{9}=2\text{ A}$. Then $R_{th}=V_{th}/I_{sc}=18/2=9\Omega$ ✅. Forecast answered: yes — with the terminals open, all 3 A is forced through $R_1$ because it is the only closed loop back to the source.

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Example 4 — Cell D: degenerate loads (short and open)

Forecast: which extreme gives the biggest current? Which gives the biggest voltage? They can't both be maxed at once — why not?

Step 1 — short circuit ($R_L=0$). The whole terminal law $V=V_{th}-I{R}_{th}$ with $V=0$: $I_{sc}=\frac{V_{th}}{R_{th}}=\frac{4}{1.33}=3\text{ A},V=0\text{ V}$ Why this step? A zero-ohm load forces the terminal voltage to zero, so all the "push" is spent across $R_{th}$ — maximum current, zero voltage.

Step 2 — open circuit ($R_L=\infty$). No current can flow: $I=0\text{ A},V=V_{th}-0\cdot R_{th}=4\text{ V}$ Why this step? Infinite resistance forbids current; with $I=0$ there's no internal drop, so the terminal shows the full $V_{th}$ — maximum voltage, zero current.

Verify: these are the two intercepts of the load line $V=4-1.33I$. At $I=0$, $V=4$ ✅; at $V=0$, $I=4/1.33=3$ ✅. Forecast answered: the short gives max current (3 A), the open gives max voltage (4 V) — they can't coincide because one always forces the other to zero. Power delivered is zero at both extremes ($P=VI$: one factor is zero each time) — which is exactly why the maximum-power load sits between them (Ex 3 of the parent).

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Example 5 — Cell E: a dependent source (test-source method)

Forecast: with no independent source, $V_{th}=0$. But $R_{th}$ need NOT be $R_1+R_2$ — the dependent source distorts it. Guess: bigger or smaller than 6 Ω?

Step 0 — confirm $V_{th}=0$. There is no independent source anywhere in this network. With the terminals left open, nothing drives any current, so $V_x=0$, which makes the dependent source produce $g\cdot 0=0$ too — the whole circuit is dead. Therefore $V_{th}=0\text{ V}.$ Why this step? A dependent source can only scale an existing signal; with no independent source to start the signal, everything sits at zero. So the only interesting number left to find is $R_{th}$.

Step 1 — why not just deactivate? A dependent source's value tracks a circuit variable ($V_x$). Killing it would erase real behaviour, so we cannot switch it off. Instead we probe: apply a test voltage $V_{test}$ at the terminals, measure the current $I_{test}$ it draws, and define $R_{th}=V_{test}/I_{test}$. Why this step? $R_{th}$ is defined as terminal voltage over terminal current — the test source measures exactly that ratio directly.

Step 2 — write the terminal current. Here $V_x=V_{test}$ (test source sits right across the terminals). Let $V_n$ be the node voltage. Current from terminal into $R_1$: $I_{test}=\frac{V_{test}-V_n}{R_1}$. At the node, the current arriving through $R_1$ leaves via two paths — down through $R_2$ to ground, and down through the dependent source (its arrow points node→ground, so $g{V}_x$ subtracts current from the node just like a resistor to ground): $\frac{V_{test}-V_n}{R_1}=\frac{V_n}{R_2}+g{V}_{test}$ Why this step? Kirchhoff's current law at the node, keeping the dependent source on with its true value $g{V}_x=g{V}_{test}$, and its downward arrow fixes the sign as a current leaving the node.

Step 3 — solve. With $R_1=4,R_2=2,g=0.5$: multiply the node equation by 4: $V_{test}-V_n=2V_n+2V_{test}\Rightarrow -V_{test}=3V_n\Rightarrow V_n=-\frac{1}{3}{V}_{test}$ Then $I_{test}=\frac{V_{test}-(-\frac{1}{3}{V}_{test})}{4}=\frac{\frac{4}{3}{V}_{test}}{4}=\frac{V_{test}}{3}$ $R_{th}=\frac{V_{test}}{I_{test}}=3\Omega$ Why this step? Substituting $V_n$ back gives $I_{test}$ purely in terms of $V_{test}$; their ratio is a pure resistance (the test voltage cancels — as it must for a linear network).

Verify: without the dependent source ($g=0$) the same algebra gives $V_n=V_{test}\cdot \frac{R_2}{R_1+R_2}$ and $R_{th}=R_1+R_2=6\Omega$. The dependent source halved it to 3 Ω. Forecast answered: smaller than 6 Ω — confirming you'd get a wrong answer if you naively wrote $R_1+R_2$. (Set $g=0$ in the code check.)

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Example 6 — Cell F: a balanced bridge, $V_{th}=0$

Forecast: the bridge looks alive (10 V feeding it). Will there be any voltage across P–Q?

Reading the figure: the two black dots at the very top and bottom are the +10 V rail and the 0 V rail. The left arm stacks orange $R_1$ over blue $R_3$, with red tap point P in the middle; the right arm stacks green $R_2$ over red $R_4$, with red tap point Q in the middle. Both taps are annotated "5 V" — that equality is the punchline. The dashed grey line joining P and Q is where you'd measure $V_{th}$; because both dots sit at the same height (same voltage), the green arrow labels the difference as zero.

Step 1 — voltage at each tap. Each arm is its own divider: $V_P=V_S\frac{R_3}{R_1+R_3}=10\cdot \frac{100}{200}=5\text{ V},V_Q=V_S\frac{R_4}{R_2+R_4}=10\cdot \frac{200}{400}=5\text{ V}$ Why this step? No load across P–Q means each arm carries its own current independently — two separate dividers.

Step 2 — the Thevenin voltage. It's the difference between the taps: $V_{th}=V_P-V_Q=5-5=0\text{ V}$ Why this step? $V_{th}$ is the open-circuit voltage across the terminals P–Q, i.e. $V_P-V_Q$. The bridge is balanced ($R_1/R_3=R_2/R_4$), so both taps sit at the same height — a live-looking circuit that delivers nothing.

Step 3 — $R_{th}$ still exists. Deactivate $V_S$ (short it). Then top rail = bottom rail = ground node. From P: $R_1\Vert R_3=100\Vert 100=50\Omega$. From Q: $R_2\Vert R_4=200\Vert 200=100\Omega$. Looking P→Q these two are in series: $R_{th}=50+100=150\Omega$ Why this step? $V_{th}=0$ does not mean $R_{th}=0$; the network still has stiffness. Shorting $V_S$ merges the rails, collapsing each arm into a parallel pair, and P-to-Q traverses both pairs in series.

Verify: balance condition $R_1{R}_4=R_2{R}_3$? $100\cdot 200=20000=200\cdot 100$ ✅ ⇒ $V_{th}=0$ guaranteed. $R_{th}=50+100=150\Omega$ ✅. Forecast answered: no — despite the live 10 V feed, the balanced bridge shows zero volts across P–Q. A short-circuit current across P–Q would be $I_{sc}=V_{th}/R_{th}=0$ — consistent with zero source voltage.

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Example 7 — Cell G: real-world word problem

Forecast: the speaker never sees a "perfect 9 V." Will $V_{th}$ be exactly 9 V? Where does the wire resistance go — into $V_{th}$ or into $R_{th}$?

Step 1 — identify the network vs. the load. The load is the $8\Omega$ speaker. Everything before it — battery + internal $r$ + wire — is the network. Why this step? Thevenin only ever describes what a two-terminal box does to whatever you attach; you must first decide where the "terminals" cut. Here you care about the speaker's experience, so the cut is at the speaker's own two wires — putting all the fixed hardware (battery, $r$, wire) on the network side and only the speaker on the load side.

Step 2 — $V_{th}$. With the speaker unplugged, no current flows, so there's zero drop across $r$ and the wire. The open-circuit voltage is the full battery EMF: $V_{th}=9\text{ V}$ Why this step? Series resistances only drop voltage when current flows; open circuit ⇒ zero current ⇒ zero drop ⇒ full EMF appears at the terminals unchanged.

Step 3 — $R_{th}$. Deactivate the battery (short its ideal EMF). Then $r$ and $R_{wire}$ are simply in series along the path to the terminals: $R_{th}=r+R_{wire}=0.5+0.3=0.8\Omega$ Why this step? All internal series resistances add up into one "stiffness" number — this is where the wire and internal resistance actually live, not in $V_{th}$.

Step 4 — current into the speaker. $I=\frac{V_{th}}{R_{th}+R_L}=\frac{9}{0.8+8}=1.02\text{ A}$ Why this step? With the network reduced to $V_{th}$ behind $R_{th}$, the speaker is just one more series resistor — a single Ohm's-law division gives the current.

Verify: voltage actually reaching the speaker $=I\cdot R_L=1.02\cdot 8=8.18\text{ V}$, so $8.18/9=91\%$ of the EMF reaches the load — the other $9\%$ is lost in $r$ and the wire. Terminal-law check: $9-1.02\cdot 0.8=8.18\text{ V}$ ✅ — the two ways of getting the speaker voltage agree. Forecast answered: $V_{th}$ is exactly 9 V (open circuit sees no drops), and the wire + internal resistance both land in $R_{th}$, not $V_{th}$.

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Example 8 — Cell H: exam twist, work backwards from measurements

Forecast: the open reading is one of the two numbers already. Which one? And does the sag from 6 V to 4 V tell you $R_{th}$ is bigger or smaller than $4\Omega$?

Step 1 — read off $V_{th}$. Open circuit means zero current means the terminal shows $V_{th}$ directly: $V_{th}=6\text{ V}$ Why this step? By definition $V_{th}$ is the open-circuit voltage — the first meter reading hands it to us with no computation.

Step 2 — use the loaded reading. With $R_L=4\Omega$, current is $I=V/R_L=4/4=1\text{ A}$. Apply the terminal law: $V=V_{th}-I{R}_{th}\Rightarrow 4=6-1\cdot R_{th}\Rightarrow R_{th}=2\Omega$ Why this step? Two measurements = two points on the load line $V=V_{th}-I{R}_{th}$; the slope between them is $-R_{th}$, so a second reading is exactly what pins down the resistance.

Verify: the sag $\Delta V=6-4=2\text{ V}$ over $I=1\text{ A}$ gives $R_{th}=\Delta V/I=2\Omega$ ✅. Cross-check: predicted loaded voltage $=V_{th}\frac{R_L}{R_{th}+R_L}=6\cdot \frac{4}{6}=4\text{ V}$ ✅. Forecast answered: the open reading is $V_{th}$; and since the load kept more than half the voltage (4 of 6 V), $R_{th}<R_L$ — here $2\Omega <4\Omega$.

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Connections

• Norton Equivalent Circuits — every $I_{sc}$ we computed is the Norton current $I_N$.
• Superposition Theorem — the engine behind Example 2's two-source add-up.
• Maximum Power Transfer Theorem — the degenerate loads of Example 4 bracket the matched-power sweet spot.
• Voltage and Current Dividers — the workhorse for $V_{th}$ in Ex 1, 6, 8.
• Source Transformation — turns Example 3's current source into a voltage source if you prefer.
• Nodal and Mesh Analysis — general method used in Example 2.